{"id":26224,"date":"2021-07-01T16:00:36","date_gmt":"2021-07-01T10:30:36","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26224"},"modified":"2021-07-01T19:19:00","modified_gmt":"2021-07-01T13:49:00","slug":"ncert-solutions-for-class-10-maths-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-5-ex-5-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1"},"content":{"rendered":"

In this Page, you will learn how to solve questions of Ex 5.1 Class 10 Maths<\/a> NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus.<\/p>\n

These NCERT Solutions for Class 10 Maths<\/a> Chapter 5 Arithmetic Progressions Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nIn which of the following situations, does the list of numbers involved make an arithmetic progression and why?
\n(i) The taxi fare after each km when the fare is \u20b9 15 for the first km and \u20b9 8 for each additional km.
\n(ii) The amount of air present in a cylinder when a vacuum pump removes \\(\\frac { 1 }{ 4 }\\) of the air remaining in the cylinder at a time.
\n(iii) The cost of digging a well after every metre of digging, when it costs \u20b9 150 for the first metre and rises by \u20b9 50 for each subsequent metre.
\n(iv) The amount of money in the account every year, when \u20b9 10000 is deposited at compound interest at 8% per annum.
\nSolution:
\n(i) Yes, 15, 23, 31, …. forms an AP as each succeeding term is obtained by adding 8 in its preceeding term.
\n(ii) No, volumes are V, \\(\\frac { 3V }{ 4 }\\), (\\(\\frac { 3V }{ 4 }\\))\u00b2, L.
\n(iii) Yes, 150,200,250,…. form an A.P.
\n(iv) No, Amounts are
\n\"NCERT
\n10000(1 + \\(\\frac { 8 }{ 100 }\\)), 10000(1 + \\(\\frac { 8 }{ 100 }\\))\u00b2, 10000(1 + \\(\\frac { 8 }{ 100 }\\))\u00b2, L<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nWrite first four terms of the AP, when the first term a and the common difference d are given as follows:
\n(i) a = 10, d = 10
\n(n) a = -2, d = 0
\n(iii) a = 4, d = -3
\n(iv) a = -1, d = \\(\\frac { 1 }{ 2 }\\)
\n(v) a = -1.25, d = -0.25
\nSolution:
\n(i) Given: a = 10, d = 10
\na1<\/sub> = 10,
\na2<\/sub> = 10 + 10 = 20
\na3<\/sub> = 20 + 10 = 30
\na4<\/sub> = 30 + 10 = 40
\nThus, the first four terms of the AP are 10, 20, 30, 40.<\/p>\n

(ii) Given: a = – 2, d = 0
\nThe first four terms of the AP are -2, -2, -2, -2.<\/p>\n

(iii) a1<\/sub> = 4, d = -3
\na2<\/sub> = a1<\/sub> + d = 4 – 3 = 1
\na3<\/sub> = a2<\/sub> + d = 1 – 3 = -2
\na4<\/sub> = a3<\/sub> + d = -2 – 3 = -5
\nThus, the first four terms of the AP are 4, 1, -2, … -5.<\/p>\n

(iv) \"NCERT<\/p>\n

(v) a1<\/sub> = -1.25, d = -0.25
\na2<\/sub> = a1<\/sub> + d = -1.25 – 0.25 = -1.50
\na3<\/sub> = a2<\/sub> + d = -1.50 – 0.25 = -1.75
\na4<\/sub> = a3<\/sub> + d = -1.75 – 0.25 = -2.00
\nThus, the first four terms of the AP are -1.25, -1.50, -1.75, -2.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFor the following APs, write the first term and the common difference:
\n(i) 3, 1, -1, -3, ……
\n(ii) -5, -1, 3, 7, ……
\n(iii) \\(\\frac { 1 }{ 3 }\\) , \\(\\frac { 5 }{ 3 }\\) , \\(\\frac { 9 }{ 3 }\\), \\(\\frac { 13 }{ 3 }\\) , ……..
\n(iv) 0.6, 1.7, 2.8, 3.9, …….
\nSolution:
\n(i) a1<\/sub> = 3, a2<\/sub> = 1
\nd = a2<\/sub> – a1<\/sub> = 1 – 3 = -2
\nwhere, a1<\/sub> = first term and d = common difference
\na1<\/sub> = 3, d = -2<\/p>\n

(ii) a1<\/sub> = -5, a2<\/sub> = -1
\nd = a2<\/sub> – a1<\/sub> = -1 – (-5) = -1 + 5 = 4
\nSo, first term a1<\/sub> = -5 and common difference d = 4<\/p>\n

(iii) a1<\/sub> = \\(\\frac { 1 }{ 3 }\\), a2<\/sub> = \\(\\frac { 5 }{ 3 }\\)
\nd = \\(\\frac { 5 }{ 3 }\\) – \\(\\frac { 1 }{ 3 }\\) = \\(\\frac { 4 }{ 3 }\\)
\nSo, first term a1<\/sub> = \\(\\frac { 1 }{ 3 }\\) and common difference d = \\(\\frac { 4 }{ 3 }\\)
\na1<\/sub>= 0.6, a2<\/sub> = 1.7
\nd = a2<\/sub> – a1<\/sub> = 1.7 – 0.6 = 1.1
\nSo, first term a1<\/sub> = 0.6 and common difference d = 1.1<\/p>\n

Question 4.
\nWhich of the following are APs ? If they form an AP, find the common difference d and write three more terms.
\n(i) 2, 4, 8, 16, …….
\n(ii) 2, \\(\\frac { 5 }{ 2 }\\) , 3, \\(\\frac { 7 }{ 2 }\\) , …….
\n(iii) -1.2, -3.2, -5.2, -7.2, ……
\n(iv) -10, -6, -2,2, …..
\n(v) 3, 3 + \u221a2, 3 + 2\u221a2, 3 + 3\u221a2, …..
\n(vi) 0.2, 0.22, 0.222, 0.2222, ……
\n(vii) 0, -4, -8, -12, …..
\n(viii) \\(\\frac { -1 }{ 2 }\\) , \\(\\frac { -1 }{ 2 }\\) , \\(\\frac { -1 }{ 2 }\\) , \\(\\frac { -1 }{ 2 }\\) , …….
\n(ix) 1, 3, 9, 27, …….
\n(x) a, 2a, 3a, 4a, …….
\n(xi) a, a2, a3, a4, …….
\n(xii) \u221a2, \u221a8, \u221a18, \u221a32, …..
\n(xiii) \u221a3, \u221a6, \u221a9, \u221a12, …..
\n(xiv) 1\u00b2, 3\u00b2, 5\u00b2, 7\u00b2, ……
\n(xv) 1\u00b2, 5\u00b2, 7\u00b2, 7\u00b2, ……
\nSolution:
\n(i) 2, 4, 8, 16, ……
\na2<\/sub> – a1<\/sub> = 4 – 2 = 2
\na3<\/sub> – a2<\/sub> = 8 – 4 = 4
\ni.e., d = an+1<\/sub> – an<\/sub> is not same every time, so the given series do not form of an A.P.<\/p>\n

(ii) We have given the series,
\ni.e., d = an+1<\/sub> – an<\/sub> an is same everything, so the given list of numbers form an AP.
\nSo, common difference (d) = \\(\\frac { 1 }{ 2 }\\)
\nNext three terms after the last given term are
\n\\(\\frac { 7 }{ 2 }\\) + \\(\\frac { 1 }{ 2 }\\) = 4
\n4 + \\(\\frac { 1 }{ 2 }\\) = \\(\\frac { 9 }{ 2 }\\) and \\(\\frac { 9 }{ 2 }\\) + \\(\\frac { 1 }{ 2 }\\) = 5
\nThus, we obtain three terms as 4, \\(\\frac { 9 }{ 2 }\\), 5.<\/p>\n

(iii) We have given the series,
\n– 1.2, – 3.2, – 5.2, – 7.2, …
\n\u2234 a2<\/sub> – a1<\/sub> = – 3.2 – (1.2) = – 2
\na3<\/sub> – a2<\/sub> = – 5.2 – (- 3.2) = – 2
\ni.e., d = an+1<\/sub> – a<sub<n is same every time, so the given list of numbers form an AP.
\nSo, common difference (d) = – 2
\nNext three terms after the last term are,
\n– 7.2 + (- 2) = – 9.2
\n– 9.2 + (- 2) = – 11.2 and – 11.2 + (- 2), – 13.2
\nThus, we obtain three terms as – 9.2, 11.2, – 13.2.<\/p>\n

(iv) We have gives series,
\n10, – 6, -2, 2
\n\u2234 a2<\/sub> – a1<\/sub> = – 6 – (- 10) = 4
\na3<\/sub> – a2<\/sub> = – 2 -( – 6) = 4
\ni.e., d = an+1<\/sub> is same every time, so the given list of numbers form an AP.
\nSo, common difference (d) = 4
\nNext three terms after after last given term
\nare,
\n2 + 4= 6
\n6 + 4 = 10 and 10 + 4 = 14 Thus, we obtain three terms as 6,10,14.<\/p>\n

\"NCERT<\/p>\n

(vi) We have given the series,
\n0.2, 0.22, 0.222, 0.2222,…
\n\u2234 a2<\/sub> – a1<\/sub> = 0.2 – 0.20 = 0.02
\na3<\/sub> – a2<\/sub> = 0.222 – 0.22 = 0.002
\ni.e., d = an+1<\/sub> is not same everytime, so the given series do not form an AP.<\/p>\n

(vii) We have given the series,
\n0, – 4, – 8, – 12, …
\n= a2<\/sub> – a1<\/sub> = – 4 – 0 = – 4
\na3<\/sub> – a2<\/sub> = – 8 – (- 4) = – 4
\ni. e., d = an+1<\/sub> – an<\/sub> is same everytime, so the given series is in the form of an AP.
\nSo, common difference (d) = – 4
\nNext three terms after the last given terms
\nare,
\n– 12 + (- 4)= – 16
\n– 16 + (- 4) = – 20 and – 20 + (- 4) = – 24
\nThus, we obtain three as – 16, – 20, – 24.<\/p>\n

(viii) We have given the series,
\n\"NCERT<\/p>\n

i.e., d = an+1<\/sub> – an<\/sub> is same everytime, i.e., 0.
\nso the given series do not form an AP.
\nNext three terms after the last given terms are
\n\\(\\frac { -1 }{ 2 }\\) + 0 = \\(\\frac { -1 }{ 2 }\\), \\(\\frac { -1 }{ 2 }\\) + 0 = \\(\\frac { -1 }{ 2 }\\) and \\(\\frac { -1 }{ 2 }\\) + 0 = \\(\\frac { -1 }{ 2 }\\)<\/p>\n

(ix) We have given the series, 1, 3, 9, 27,…
\n\u2234 a2<\/sub> – a1<\/sub> = 3 – 1 = 2
\na3<\/sub> – a2<\/sub> = 9 – 3 = 6
\ni.e., d = an+1<\/sub> – an<\/sub> is not always same.
\nSo, the given series are, do not form an AP.<\/p>\n

(x) We have given the series, a, 2a, 3a, 4a, …
\n\u2234 a2<\/sub> – a1<\/sub> = 2a – a = a
\na3<\/sub> – a2<\/sub> = 3a – 2a = a
\ni.e., d = an+1<\/sub> – an<\/sub> is same everytime, so the given series is in the form of an A.P.
\nSo, common difference (d) = a
\nNext three terms after the last given term are, 4a + a = 5a
\n5a + a = 6a and 6a + a = 7a Thus, we obtain three terms as 5a, 6a, 7a.<\/p>\n

(xi) We have given the series,
\na1<\/sub>, a\u00b2, a\u00b3, a4<\/sup>, ….
\n\u2234 a\u00b22<\/sub> – a1<\/sub> = a (a – 1)
\na\u00b3 – a\u00b2 = a\u00b2 (a – 1)
\ni.e., d = an+1<\/sub> – an<\/sub> is not always same.
\nSo the given series are do not an A.P.<\/p>\n

(xii) We have given the series,
\n\"NCERT
\ni.e., d = an+1<\/sub> – an<\/sub> is same everytime so, the given series is in the form of an AP.
\nSo, common difference (d) = \\(\\sqrt{2}\\)
\nNext three terms after the last terms ate,
\n\"NCERT
\nThus, we obtain three terms as
\n\\(\\sqrt{50}\\), \\(\\sqrt{72}\\), \\(\\sqrt{98}\\).<\/p>\n

(xiii) We have gives the series
\n\"NCERT
\ni.e., d = an+1<\/sub> – an<\/sub> is not same evervtime, so the given sereis are do not form an AP.<\/p>\n

(xiv) We have given the series, 1\u00b2, 3\u00b2, 5\u00b2, 7\u00b2
\n\u2234 a2<\/sub> – a1<\/sub> = 3\u00b2 – 1\u00b2 = 9 – 1 = 8
\na3<\/sub> – a2<\/sub> = 5\u00b2 – 32 = 25 – 9 = 16
\ni.e., d = an+1<\/sub> – an<\/sub> is not same everytime, so the given series do not form an AP.<\/p>\n

(xv) We have given the series, 1\u00b2, 5\u00b2, 7\u00b2, 7\u00b2, …
\n\u2234 a2<\/sub> – a1<\/sub> = 5\u00b2 – 1\u00b2 = 25 – 1 = 24
\na3<\/sub> – a2<\/sub> = 7\u00b2 – 5\u00b2 = 49 – 25 = 24
\na4<\/sub> – a3<\/sub> = 7\u00b3 – 7\u00b2 = 73 – 49 = 24
\ni.e., d = an+1<\/sub> – an <\/sub>is same everytime, so the given series are in the form of an AP.
\nSo, common difference (d) = 24
\nNext three terms after the last term are,
\n73 + 24 = 97
\n97 + 24 = 121 and 121 + 24 = 145
\nThus we obtain three terms as 97, 121, 145.<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

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